hdu 5154 -- Harry and Magical Computer

 

Harry and Magical Computer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 641    Accepted Submission(s): 295

Problem Description

In reward of being yearly outstanding magic student, Harry gets a magical computer. When the computer begins to deal with a process, it will work until the ending of the processes. One day the computer got n processes to deal with. We number the processes from 1 to n. However there are some dependencies between some processes. When there exists a dependencies (a, b), it means process b must be finished before process a. By knowing all the m dependencies, Harry wants to know if the computer can finish all the n processes.

 

Input

There are several test cases, you should process to the end of file.
For each test case, there are two numbers n m on the first line, indicates the number processes and the number of dependencies. 1≤n≤100,1≤m≤10000
The next following m lines, each line contains two numbers a b, indicates a dependencies (a, b). 1≤a,b≤n

 

Output

Output one line for each test case. 
If the computer can finish all the process print "YES" (Without quotes).
Else print "NO" (Without quotes).

 

Sample Input

3 2 3 1 2 1 3 3 3 2 2 1 1 3

 

Sample Output

YES NO

 

---

思路: 用拓扑排序求解.

 

/*======================================================================
 *           Author :   kevin
 *            Email :   zerocode.kevin@gmail.com
 *         Filename :   HarryAndMagicalComputer.cpp
 *       Creat time :   2015-01-23 09:17
 *      Description :
========================================================================*/
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#define clr(a,b) memset(a,b,sizeof(a))
#define INF 0x7f7f7f7f
#define M 10005
using namespace std;
inline int min_32(int (a),int (b)){return (a)<(b)?(a):(b);}
inline int max_32(int (a),int (b)){return (a)>(b)?(a):(b);}
inline long long min_64(long long (a),long long (b)){return (a)<(b)?(a):(b);}
inline long long max_64(long long (a),long long (b)){return (a)>(b)?(a):(b);}
int head[M],in[M];
struct Node{
    int to,next;
}node[M];
void addEdges(int i,int j,int k)
{
    node[k].to = j;
    node[k].next = head[i];
    head[i] = k;
}
int slove(int n)
{
    queue<int>que;
    for(int i = 1; i <= n; i++){
        if(in[i] == 0){
            que.push(i);
        }
    }
    int k,cnt = 0;
    while(!que.empty()){
        int t = que.front();
        que.pop();
        cnt++;
        for(k = head[t]; k != -1; k = node[k].next){
            if(--in[node[k].to] == 0){
                que.push(node[k].to);
            }
        }
    }
    if(cnt == n){
        return 1;
    }
    return 0;
}
int main(int argc,char *argv[])
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        clr(head,-1);
        clr(in,0);
        int a,b;
        for(int i = 0; i < m; i++){
            scanf("%d%d",&a,&b);
            addEdges(b,a,i);
            in[a]++;
        }
        if(slove(n)){
            printf("YES\n");
        }
        else{
            printf("NO\n");
        }
    }
    return 0;
}